Question

2. Find the equations of the straight lines passing through the point (1, 2) and making an angle of
60° with the line root3 x + y + 2 = 0.​

Answer 1

★Given :

• A straight line is passing through the point (1, 2) and making an angle of  60° with the line √3x + y + 2 = 0.​

★To find :

• Equation of the straight line.

★Solution :

Given line √3x + y + 2 = 0

It can be written as,

→ y = -√3x - 2

Therefore,

→Slope = -√3

(As it is in the form of y = mx+c.

Where 'm' is the slope.)

Let the slope of other line = m

Using the formula,

✦Tanθ = |m₁-m₂/1+m₁m₂|

Putting values,

→Tan 60° = |(m - (-√3)) / (1  + m(  -√3))|

→√3 = | (m + √3) /(1 - m√3) |

Squaring on both sides,

→3 = (m² + 3 + 2√3m) / (1 + 3m² - 2m√3)

→3 + 9m² - 6m√3 = m² + 3 + 2√3m

→8m² -  8m√3 = 0

→8m(m - √3) = 0

→8m = 0 (or) m-√3 = 0

→m = 0 or m = √3

Now,

✦As y = mx + c,

Putting m = 0 :

→y = mx + c

→y = 0(x) = c

→y = c  

Putting m = √3 :

→y = mx + c

→y = √3x + c

Given that the line passed through (1 , 2)

Putting values,

→y = 2    or   2 = √3 + c  

→y = 2 or  c = 2 - √3

→ y = 2 or y = √3x +  2 - √3

→y = 2 or y=√3 (x-1)+2

Hence,

The equations of the straight line is y = 2 or y=√3 (x-1)+2.

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